Question: The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $11.4$ years; the standard deviation is $2.8$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living less than $19.8$ years.
Solution: $11.4$ $8.6$ $14.2$ $5.8$ $17$ $3$ $19.8$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $11.4$ years. We know the standard deviation is $2.8$ years, so one standard deviation below the mean is $8.6$ years and one standard deviation above the mean is $14.2$ years. Two standard deviations below the mean is $5.8$ years and two standard deviations above the mean is $17$ years. Three standard deviations below the mean is $3$ years and three standard deviations above the mean is $19.8$ years. We are interested in the probability of a meerkat living less than $19.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the meerkats will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the meerkats will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $3$ years and the other half $({0.15\%})$ will live longer than $19.8$ years. The probability of a particular meerkat living less than $19.8$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.